$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Sp{Sp}$
Last time, I asked a question on the computation about modulus character of parabolic subgroup of symplectic group and LSpice gave me a nice answer for it using positive root associated to parabolic subgroup. (See Question on the modulus character of classical p-adic group. )
In the meanwhile, I have a question on the computation on a subgroup of symplectic group, which is an almost parabolic subgroup but not parabolic group. I am wondering whether we can compute it using the roots associated to the root groups appearing in the unipotent subgroup of the 'almost' parabolic subgroup.
Let me give a more explanation on this.
Let $\Sp(2n)$ be the isometry group of some $2n$-dimensional symplectic space $W$ over a $p$-adic local field $F$. Let $P_k=M_k U_k$ is a standard parabolic subgroup of $\Sp(2n)$ whose Levi subgroup $M_k$ is isomorphic to $\GL_k \times \Sp(2n-2k)$ and $U_k$ is the unipotent radical of $P_k$.
Let $N_n$ be the unipotent radical of the Borel subgroup of $\GL_n$.(i.e. upper triangular matrices with diagonal entries are 1.) Let $P_{n,k}=M_{n,k}U_{n,k}$ be the parabolic subgroup of $\GL_n$, where $M_{n,k}=GL_k \times \GL_{n-k}$ and $U_{n,k}$ is the unipotent radical of $P_{n,k}$. Consider the subgroup $R_{n,k}=(\GL_k \times N_{n-k}) \rtimes U_{n,k}$ of $\GL_n$.
We also consider an 'almost' parabolic subgroup $T_{n,k}=R_{n,k} \rtimes U_n \subset \GL_n \rtimes U_n$ of $\Sp(2n)$. Then I am wondering the modulus character of $T_{n,k}$ restricted to $\GL_{k}$.
Though $T_{n,k}$ is not parabolic subgroup of $\Sp(2n)$, it has the unipotent radical $\big((1 \times N_{n-k}) \rtimes U_{n,k}\big) \rtimes U_n$. Therefore, the positive roots corresponding this unipotent radical is the set $\{\epsilon_{i}\}_{1\le i \le n} \bigcup \{ \epsilon_{i}+\epsilon_{j}\}_{1\le i < j \le n}\bigcup \{ \epsilon_{i}-\epsilon_{j}\}_{1\le i \le k <j \le n }\bigcup \{ \epsilon_{i}-\epsilon_{j}\}_{k< i <j \le n }$.
Therefore, if we restrict the modulus character $\delta_{T_{n,k}}$ to $\GL_k$, then it is $(2n-k+1)(\epsilon_1+\cdots +\epsilon_k)$.
Therefore, I guess that $\delta_{T_{n,k}}|_{\GL_k}=|\det_{\GL_k}|^{2n-k+1}$.
Is this right?
Any comments are highly appreciated!
